Question

Please help me solve this unit operations problem.(Fundamentals of momentum, heat and mass transfer)

For straight operating and equilibrium lines, how to evaluate the proper average driving force for the transfer process by those at the two ends of the column? (5% ) Show how the design of the equipment can be more appropriate by considering the fact that the equilibrium line is slightly curved. (3%)

Answer 1

The figure is a mass transfer process whose equilibrium line and the operating line is a straight line.

Operating Line Equilibrium (bottom) Line Ул P (x, y У top) Y2 mole fraction solute in liqu Х х х mole friction solute in vapour, y

Moving up the column the mole fraction of solute in liquid as well as decreases. The mole fraction of solute in both the phases decreases as we go up the column.

The magnitude of distance between the equilibrium line and operating line decreases is the magnitude of the driving force.

At the bottom of the column

For liquid phase

Average Mole fraction driving force = x₁^* - x₁

For gas phase

Average mole fraction driving force = y₁ - y₁^*

At top of the column

For liquid phase

Average mole fraction driving force = x₂^* - x₂

For gas phase

Average mole fraction driving force = y₂ - y₂*

The equilibrium curve and the operating curve is a straight line when we assume that the molar flow rate of liquid and gas is approximately constant.

Actual flow rate changes due to the transfer process. Transfer processes involve the movement of particles from one phase to another. Assuming the flow rate to be constant is helpful is simpler calculations but the actual flow rate gives the most accurate answer.

Slightly curved line approaches the actual values of mole fractions of the different phases.